A random sample of size 26 is to be selected from a population that has a mean ì = 46 and a standard deviation ó of 15.
(a) This sample of 26 has a mean value of x, which belongs to a sampling distribution. Find the shape of this sampling distribution.
skewed right approximately normal skewed left chi-square
(b) Find the mean of this sampling distribution. (Give your answer correct to nearest whole number.) 46
(c) Find the standard error of this sampling distribution. (Give your answer correct to two decimal places.)
2.94
(d) What is the probability that this sample mean will be between 40 and 55? (Give your answer correct to four decimal places.)
.9782
(e) What is the probability that the sample mean will have a value greater than 52? (Give your answer correct to four decimal places..0207
(f) What is the probability that the sample mean will be within 6 units of the mean? (Give your answer correct to four decimal places.)
Have no ideal worked this one as I have the previous ones and it is wrong
To find the probability that the sample mean will be within 6 units of the mean, we'll need to use the concept of the standard error.
The standard error, denoted as SE, measures the average amount that the sample mean differs from the population mean. It is calculated using the formula:
SE = σ / √n
where σ is the population standard deviation, and n is the sample size.
In this case, the population standard deviation (σ) is given as 15, and the sample size (n) is 26. Plugging these values into the formula, we can calculate the standard error:
SE = 15 / √26 ≈ 2.94 (rounded to two decimal places)
Now, to find the probability that the sample mean will be within 6 units of the mean, we need to find the corresponding z-scores for the lower and upper limits.
Lower limit z-score:
Z = (X - μ) / SE
Z = (40 - 46) / 2.94 ≈ -2.04
Upper limit z-score:
Z = (X - μ) / SE
Z = (55 - 46) / 2.94 ≈ 3.06
To find the probability between these z-scores, we can use a standard normal distribution table or a calculator.
From the standard normal distribution table or calculator, the probability corresponding to a z-score of -2.04 is approximately 0.0207, and the probability corresponding to a z-score of 3.06 is approximately 0.9782.
To find the probability between the two limits, we subtract the lower probability from 1 (to include the area up to the lower limit) and then subtract the upper probability from the result:
P(40 ≤ X ≤ 55) ≈ 1 - 0.0207 - 0.9782 ≈ 0.0011
Therefore, the probability that the sample mean will be within 6 units of the mean is approximately 0.0011 (correct to four decimal places).